Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 2 - Functions and Their Graphs - Chapter Review - Review Exercises - Page 115: 24

Answer

$\text{(a)}$ $23$. $\text{(b)}$ $7$. $\text{(c)}$ $47$

Work Step by Step

Given $f(x)=8x^2-x$, the average rate of change $R$ is given by the formula $$R=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ Thus, $\text{(a)}$ For $x_1=1, x_2=2$, we have: $R=\dfrac{f(2)-f(1)}{2-1}=\dfrac{8(2)^2-(2)-[8(1)^2-(1)]}{1}=23$. $\text{(b)}$ For $x_1=0, x_2=1$, we have: $R=\dfrac{f(1)-f(0)}{1-0}=\dfrac{8(1)^2-(1)-[8(0)^2-(0)]}{1}=7$. $\text{(c)}$ For $x_1=2, x_2=4$, we have: $R=\dfrac{f(4)-f(2)}{4-2}=\dfrac{8(4)^2-(4)-[8(2)^2-(2)]}{2}=47$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.